Q183 / 270 · NDA 2017 (Phase 2) · Mathematics

The value of $\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}$ is equal to

  • Correct answer
    A
    $4$
  • B
    $2$
  • C
    $1$
  • D
    $-4$

Solution

$\begin{aligned} & \sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}=\frac{\sqrt{3}}{\sin 20^{\circ}}-\frac{1}{\cos 20^{\circ}} \\ & =\frac{\sqrt{3} \cos 20^{\circ}-\sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}\end{aligned}$ $\begin{aligned} & =\frac{2\left(\frac{\sqrt{3}}{2} \cos 20^{\circ}-\frac{1}{2} \sin 20^{\circ}\right)}{\sin 20^{\circ} \cos 20^{\circ}} \\ & =\frac{2\left(\cos 30^{\circ} \cos 20^{\circ}-\sin 30^{\circ} \sin 20^{\circ}\right)}{\left(\frac{2 \sin 20^{\circ} \cos 20^{\circ}}{2}\right)}\end{aligned}$ $\begin{aligned} & =\frac{4\left[\cos \left(30^{\circ}+20^{\circ}\right)\right]}{\sin 40^{\circ}} \\ & =\frac{4 \cos 50^{\circ}}{\sin 40^{\circ}}=\frac{4 \cos \left(90^{\circ}-40^{\circ}\right)}{\sin 40^{\circ}}=\frac{4 \sin 40^{\circ}}{\sin 40^{\circ}}=4 .\end{aligned}$

Hence, the correct answer is option: 'A'.

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